Mobius function and Euler Product Triangle Gevorg Hmayakyan proposed last year (17 September 2017) on StackExchange the following relationship concerning the mobius function. Gevorg stated that if
$$∏^{n−1}_{i=1}(1−x^i)=∑_{k=0}^{n(n−1)/2}a_{n,k}x^{k} $$
then
$$μ(n)=a_{n,1}+a_{n,n+1}+a_{n,2n+1}+a_{n,3n+1}+...$$
Does anyone know how prove this?
During related computer computation on this function, I also noted that
$$\phi (n)=a_{n,0}+a_{n,n}+a_{n,2n}+a_{n,3n}+...$$
but why this is the case I do not know? Any insights?
PS. I had noticed that the above generating function is equivalent to a row of Euler Truncated Product Triangle (see the paper by Alex Mennen at http://www.alex.mennen.org/mahoniantri.pdf). Thus
Using Mennens notation then they become
$$μ(n)=∑_{k=0}^{n(n−1)/2}P(n−2,kn+1)$$
$$\phi(n)=∑_{k=0}^{n(n−1)/2}P(n−2,kn)$$
 A: For $\phi(n)$, series multisection means that
$$\sum_{k}a_{n,kn}x^{kn}=\frac1n \sum_{j=0}^n f_n(\zeta^j x)$$
where $f_n(x)=\prod_{i=1}^{n-1}(1-x^i)$ and $\zeta=\exp(2\pi i/n)$. Therefore
$$\sum_{k}a_{n,kn}=\frac1n \sum_{j=0}^n f_n(\zeta^j).$$
If $\gcd(j,n)>1$ then $f_n(\zeta^j)=0$. For $\gcd(j,n)=1$ then
$f_n(\zeta^j)=f_n(\zeta)=\prod_{j=1}^{n-1}(1-\zeta_j)$ and so
$$\sum_{k}a_{n,kn}=\frac{\phi(n)}n f_n(\zeta).$$
But
$$f_n(\zeta)=\lim_{x\to1}\frac{(x-1)(x-\zeta)(x-\zeta^2)\cdots(x-\zeta^{n-1})}{x-1}=\lim_{x\to1}\frac{x^n-1}{x-1}=n.$$
We conclude $$\sum_{k}a_{n,kn}=\phi(n).$$
ADDED IN EDIT
The $\mu(n)$ case follows similar lines. In this case
$$\sum_{k}a_{n,kn+1}=\frac1n \sum_{j=0}^n\zeta^{-j} f_n(\zeta^j).$$
This reduces to $\sum_{j:gcd(n,j)=1}\zeta^{-j}$. This is the
sum of all primitive $n$-th roots of unity, which is $\mu(n)$.
A: Thank you "Lord Shark the Unknown". What an elegant answer. Since then I have done some further research based on your answer in relation to the general case. I deduced from Wikipedia that these are related to Ramujanim Sums.
$$\sum_{k=0}^{n(n-1)/2}a_{n,kn+j} = c_n(j) =\mu (\frac{n}{(n,j)})\frac{\phi (n)}{\phi (\frac{n}{n,j})}$$
